The string “PAYPALISHIRING” is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
P A H N A P L S I I G Y I R
And then read line by line: “PAHNAPLSIIGYIR”
Write the code that will take a string and make this conversion given a number of rows:
string convert(string s, int numRows);
Example 1:
Input: s = "PAYPALISHIRING", numRows = 3 Output: "PAHNAPLSIIGYIR"
Example 2:
Input: s = "PAYPALISHIRING", numRows = 4 Output: "PINALSIGYAHRPI" Explanation: P I N A L S I G Y A H R P I
将字符串 “PAYPALISHIRING” 以Z字形排列成给定的行数:
P A H N A P L S I I G Y I R
之后从左往右,逐行读取字符:”PAHNAPLSIIGYIR”
实现一个将字符串进行指定行数变换的函数:
string convert(string s, int numRows);
示例 1:
输入: s = "PAYPALISHIRING", numRows = 3 输出: "PAHNAPLSIIGYIR"
示例 2:
输入: s = "PAYPALISHIRING", numRows = 4 输出: "PINALSIGYAHRPI" 解释: P I N A L S I G Y A H R P I
纯模拟题,简单找一下规律就可以了,然后直接输出就行,代码如下:
char* convert(char* s, int numRows) { int len = strlen(s); char *sres = calloc(1, len + 1); int i = 0; int j = 0; int pos = 0; int gap = (numRows - 1) * 2; int step = 0; if (numRows <= 1) return s; for (i = 0; i < numRows; i++) { step = 2 * (numRows - i % (numRows - 1) - 1); for (j = i; j < len;) { sres[pos++] = s[j]; j += step; step = gap - step; if (step == 0) step = gap - step; } } return sres; }本文标题:[leetcode题解]6. ZigZag Conversion
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